# ### Author Question: solving a quadratic equation? (Read 1690 times)

#### j_sun

• Sr. Member
•    • Posts: 384
on: Jun 18, 2013
I was absent from school 3 days last week (for school related functions) and I missed a lot of information in my math class.  I have worked and worked at this problem for days and have not been able to figure it out.  someone please help me!

here is the question:
A rancher has 1000 feet of fencing to construct six corrals.  Find the dimensions that maximize the enclosed area.  What is the maximum area?
Solve the problem using the concepts presented this week: quadratics and the vertex maximun point on a parabola.

#### Melanie

• Full Member
•   • Posts: 204
Reply #1 on: Jun 18, 2013
Express the area in terms of the length and breadth of the [assumed rectangular] corrals, which should be a quadratic expression.
Quadratic expression has either a maximum or minimum value. In this case we expect a maximum value.
To find this value, one of the way is to by completing the square.
I doubt differentiation is what you have missed out, and I guess you haven't learned yet.

Here, if we assume the 6 corrals are separated, then need to argue that the maximum area would be achieve by constructing the maximum area for each corral using 1/6 of the available fencing.

Or, we can also assume a topology of the 6 corrals to be adjacent to one another in a row, which is more logical/natural and efficient use of the fencing.

Using the latter assumption, let x feet be the length of the row of 6 corrals. The width can be expressed in terms of x by considering 1000-2x = 7 widths, thus the width is (1000-2x)/7.

Let A squared feet be the area of the 6 corrals.
A = x * (1000-2x)/7
= 2(x^2 - 500x)/7
By completing the square:
A = -2/7 * [(x - 250)^2 - 250^2]
Thus, maximum A when (x-250)^2 is minimum, i.e. zero.
length, x = 250 feet, width = (1000-500)/7 feet
and area, A = -2/7 * (- 250^2)

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